krodkiewski dynamic book

436-202 MECHANICS 1UNIT 2

DYNAMICSOF

MACHINES

J.M. KRODKIEWSKI

2008

THE UNIVERSITY OF MELBOURNEDepartment of Mechanical and Manufacturing Engineering

1

2.

DYNAMICS OF MACHINES

Copyright C 2008 by J.M. Krodkiewski

ISBN 0-7325-1535-1

The University of MelbourneDepartment of Mechanical and Manufacturing Engineering

CONTENTS

1 DYNAMICS OF PARTICLES 51.1 KINEMATICS OF A PARTICLE . . . . . . . . . . . . . . . . . . . . 5

1.1.1 Absolute linear velocity and absolute linear acceleration 51.1.2 Intrinsic system of coordinates. . . . . . . . . . . . . . . . 61.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2 KINETICS OF PARTICLE . . . . . . . . . . . . . . . . . . . . . . . 231.2.1 Equations of motion . . . . . . . . . . . . . . . . . . . . . . 231.2.2 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.2.3 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . 251.2.4 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.2.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2 PLANE DYNAMICS OF A RIGID BODY 592.1 PLANE KINEMATICS OF A RIGID BODY . . . . . . . . . . . . . . 59

2.1.1 Non-inertial (moving) systems of coordinates . . . . . . 60

2.1.2 Motion of rigid body . . . . . . . . . . . . . . . . . . . . . 682.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

2.2 PLANE KINETICS OF A RIGID BODY . . . . . . . . . . . . . . . . 862.2.1 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . 862.2.2 Properties of the mass moment of inertia . . . . . . . . 892.2.3 Equations of motion . . . . . . . . . . . . . . . . . . . . . . 912.2.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

3 DYNAMICS OF PLANE MECHANISMS 1303.1 CONSTRAINTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1303.2 MOBILITY - GENERALIZED COORDINATES. . . . . . . . . . . . 1333.3 NUMBER OF DEGREE OF FREEDOM - DRIVING FORCES. . . . 135

3.3.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1363.4 EQUATIONS OF MOTION. . . . . . . . . . . . . . . . . . . . . . . . 139

3.4.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

4 APPENDIXES 1724.1 APPENDIX 1. REVISION OF THE VECTOR CALCULUS 172

4.2 APPENDIX 2. CENTRE OF GRAVITY, VOLUME ANDMOMENTS OF INERTIA OF RIGID BODIES. . . . . . . . . 175

CONTENTS 4

INTRODUCTION.The purpose of this text is to provide the students with the theoretical back-

ground and engineering applications of the two dimensional mechanics of the rigidbody. It is divided into four chapters.

The first one, Dynamic of particles, deals with these engineering problemsthat can be modelled by means of limited number of particles. A large class ofengineering problems falls into this category.

The second chapter, Plane dynamics of a rigid body, consists of two sec-tions. The first one, Kinematics, deals with the geometry of motion of rigid bodiesin terms of the inertial as well as in terms of the non-inertial system of coordinates.In the second section, entitled Kinetics, the equations that determine the relation-ship between motion and the forces acting on a rigid body are derived. This chapterprovides bases for development of methods of modelling and analysis of the planemechanical systems. Mechanical system usually refers to a number of rigid bodiesconnected to each other by means of the kinematic constraints or the elastic ele-ments. The last two chapters are devoted to the engineering problems that can beapproximated by the plane mechanical systems.

The third chapter, Dynamic Analysis of Plane Mechanisms, provides asystematic approach to the dynamics of a number of rigid bodies interconnected bythe kinematic constraints. Formulation of the dierential equations of motion as wellas the determination of the interaction forces in the kinematic joints are covered.

Each chapter is supplied with several engineering problems. Solution to someof them are provided. Solution to the other problems should be produced by studentsduring tutorials and in their own time.

Chapter 1

DYNAMICS OF PARTICLES

1.1 KINEMATICS OF A PARTICLE

To consider motion of a particle we assume the existence of so-called absolute (mo-tionless) system of coordinates XaYaZa(see Fig. 1).

DEFINITION: Inertial system of coordinated is one that does not rotate andwhich origin is fixed in the absolute space or moves along straight line ata constant velocity.

XY

Z

O

Oa Ya

Za

Xa

v

Figure 1

The inertial systems of coordinates are usually denoted by upper characters,e.g. XY Z, to distinguish them from non-inertial systems of coordinates that areusually denoted by lower characters, e.g. xyz.1.1.1 Absolute linear velocity and absolute linear accelerationLet us assume that motion of a particle is given by a set of parametric equations 1.1that determines the particles coordinates at any instant of time (see Fig. 2).

rX = rX(t)

rY = rY (t)

rZ = rZ(t) (1.1)

These coordinates represent scalar magnitude of components of so called absoluteposition vector r along the inertial system of coordinates XY Z.

r = IrX(t) + JrY (t) +KrZ(t) (1.2)

KINEMATICS OF A PARTICLE 6

X

Y

Z

O

K

IJ

r(t) r

(t+ t)rrZ(t)

rY(t) rX (t)

Figure 2

where I,J,K are unit vectors of the inertial system of coordinates XY Z. Vector ofthe absolute velocity, as the first derivative of the absolute position vector r withrespect to time, is given by the following formula.

v = r = limtO

rt = I limtO

rXt + J limtO

rYt +K limtO

rZt = IrX + JrY +KrZ

(1.3)Similarly, vector of the absolute acceleration is defined as the second derivative of theposition vector with respect to time.

a = r = limtO

vt = IrX + JrY +KrZ (1.4)

Scalar magnitude of velocity v (speed) can be expressed by the following formula

v = |v| = v v =qr2X + r

2Y + r

2Z (1.5)

Scalar magnitude of acceleration is

a = |a| = a a =qr2X + r

2Y + r

2Z (1.6)

The distance done by the particle in a certain interval of time 0 < < t is given bythe formula 1.7.

s =Z t0

v d =Z t0

prX()2 + rY ()2 + rZ()2d = s(t) (1.7)

1.1.2 Intrinsic system of coordinates.The formula 1.7 assigns to any instant of time a scalar distance s done by the point Pmeasured along the path from a datum usually corresponding to time t = 0 (Fig. 3).Therefore, it is possible to express the position vector r as a function of the distances.

r = IrX(s) + JrY (s) +KrZ(s) (1.8)


X

Y

Z

O

s

t=0 tt v1

r(s)

Figure 3

The vector of velocity v is tangential to the path. Unit vector t1 that possessesthe direction and sense of v is called tangential unit vector.

t1 =v

v=drdt1

v=drdtdtds=drds

(1.9)

The infinitesimal vector increment dt1 (see Fig. 4.) is normal to the vector t1 anddefines the direction and sense of the normal unit vector n1. Hence

n1 =dt1dtdt1dt

(1.10)

XY

Z

O

r(s) n1

r(s+ds)

1t (s)

1t (s+ds)

1t (s+ds)

d t1

Figure 4


XY

Z

O

r(s) n1

1t (s)

1t (s+ds)

C

T

T

1

P1

2

P2

ds

dt1

Figure 5

Vectors t1 and n1 define so-called osculating plane which contains the oscu-lating circle of radius . Since triangle CP1P2 is similar to P1T1T2 (see Fig. 5)

dt1t1=ds

(1.11)

Hence, according to Eqs.. 1.10, 1.11 and 1.9 one can obtain

n1 =dt1dt1

=dt1ds t1

=dt1ds

= dds

drds

=

d2rds2

(1.12)

Since n1 is an unit vector,

n1 =

d2rXds2

2+

d2rYds2

2+

d2rZds2

2!1/2= 1 (1.13)

Hence =

1rd2rXds2

2+d2rYds2

2+d2rZds2

2 (1.14)The third unit vector, binormal b1, which together with t1 and n1 forms so-calledintrinsic system of coordinates is defined by Eq. 1.15

b1 = t1 n1 (1.15)

The components of the vector of acceleration a along the intrinsic system of coordi-nates are

a =dvdt=

ddt(vt1) = vt1 + vt1 (1.16)

Since according to Eqs.. 1.10 and 1.11

t1 =dt1dtn1 =

ds t1 dt

n1 =1

dsdtn1 =

1

v n1 (1.17)


the expression for acceleration has the following form

a = vt1 +1

v2n1 (1.18)

The first term represents the tangential component of the absolute acceleration whereasthe second term is called the normal component of the absolute acceleration.


1.1.3 ProblemsProblem 1

Motion of a particle with respect to the inertial space is given by the followingposition vector

r = Ia cost+ Ja sint+Kbt (1.19)

Determine unit vectors that are tangential (t1), normal (n1) and binormal (b1) totrajectory of motion of the particle as well as components of its absolute accelerationalong those unit vectors.


SolutionThe equation 1.19 represents motion along the screw line coiled on the cylinder

of radius a as shown in Fig. 6

Y

screw line

X

r

Z

r Y r Z r X

O a

s

H P

Figure 6

Its parametric equations are

rX = a cost

rY = a sint

rZ = bt (1.20)

When the point P follows the screw line, its projection on the plane XY follows thecircle of the radius a.

rX = a cost

rY = a sint (1.21)

The time needed to cover the full circle is

T =2

(1.22)

The distance H that corresponds to this time (t = T ) is called lead. From the thirdequation of the set 1.20 one can see that

H = bT =2b

(1.23)


The tangential unit vector coincides the vector of the absolute velocity v.

t1 =v

v(1.24)

The absolute velocity v can be obtained by dierentiation of the absolute positionvector r given by the equation 1.19

v = r = I( a sint) + J(a cost) +K(b) (1.25)

Its length is

v = |v| =p( a sint)2 + (a cost)2 + b2 =

a22 + b2 (1.26)

Introducing equation 1.25 and 1.26 into equation 1.24 one can get

t1 =1

a22 + b2(I( a sint) + J(a cost) +K(b)) (1.27)

By definition the normal vector is

n1 =dt1dtdt1dt

= (1.28)Dierentiating the vector 1.27 with respect to time we have

dt1dt=

1a22 + b2

I( a2 cost) + J(a2 sint) +K(0)

(1.29)

Hencedt1dt

=

r1

a22 + b2(( a2 cost)2 + (a2 sint)2) = a

2

a22 + b2

(1.30)

Introduction of 1.29 and 1.30 into 1.28 yields

n1 =1

a2I( a2 cost) + J(a2 sint) +K(0)

= I( cost) + J( sint) +K(0) (1.31)

The binormal vector as the vector product of t1 and n1 is

b1 = t1n1 = 1a22 + b2

I J Ka sint a cost b cost sint 0

=1

a22 + b2(I(b sint) + J(b cost) +K(a)) (1.32)

The geometrical interpretation of these unit vectors is given in Fig. 7.


Y

osculating plane

X

r

Z

O

a

vb1

t1n1

s

Figure 7

Dierentiation of the vector of the absolute velocity yields the absolute accel-eration.

a = v = IvX + JvY +KvZ = I( a2 cost) + J(a2 sint) +K(0) (1.33)

Its components along the intrinsic system of coordinates are

at1 = t1 aan1 = n1 aab1 = b1 a (1.34)


Problem 2

The parabolic slide 1 (see Fig. 8) is motionless with respect to the inertialspace. The pin 3 follows this parabolic slide and it is driven by the slide 2. The slide2 moves with the constant velocity w. Produce

1. the expression for the absolute velocity of the pin 3 as a function of time2. the expression for the normal and tangential component of the absolute

acceleration of the pin as a function of time.

H

1

2

L

w

3

Figure 8


Solution

H

L

wt

3

Y

X

rXrYr

Figure 9

The equation of the parabolic slide is

Y = bX2 (1.35)

Since this parabola goes through point (L,H)

H = bL2 (1.36)

Henceb =

HL2

(1.37)

The displacement along the X coordinate is wt. Therefore the components of theposition vector r (see Fig. 9) are

r = IrX + JrY = Iwt+ Jbw2t2 (1.38)

The absolute velocity of the pin, as the first derivative of the absolute position vectorwith respect to time, is

v = r = Iw + J2bw2t (1.39)

Its magnitude is

|v| = v =w2 + 4b2w4t2 = w

1 + 4b2w2t2 (1.40)

The tangential unit vector is

t1 =v

|v| =1

1 + 4b2w2t2(I+ J2bwt) (1.41)


The normal vector is equal to

n1 =dt1dtdt1dt

(1.42)where

dt1dt=

1

(1 + 4b2w2t2)32

I4b2w2t

+ J (2bw)

(1.43)

and dt1dt

=

2bw1 + 4b2w2t2

(1.44)

Hencen1 =

11 + 4b2w2t2

(I (2bwt) + J (1)) (1.45)

The absolute acceleration according to 1.39is

a = v = I(0) + J2bw2 (1.46)

Its tangential component is

at1 = t1 a =

11 + 4b2w2t2

(I+ J2bwt) I(0) + J2bw2 =

=4b2w3t1 + 4b2w2t2

(1.47)

The normal component as the dot product of the normal vector n1 and the accelera-tion a is

an1 = n1 a =

11 + 4b2w2t2

(I (2bwt) + J (1)) I(0) + J2bw2 =

=2bw2

1 + 4b2w2t2(1.48)


Problem 3

y1

x1

Y

X

z1 Z

x1

O

O

= t

= t

1 2

1 2

R

Figure 10

The circular slide 1 of radius R (see Fig. 10) rotates about the vertical axisZ of the inertial system of coordinates XY Z. Its rotation is determined by the angle = t. The system of coordinates x1y1z1 is attached to the slide 1. The element 2can be treated as a particle. The relative motion of the particle 2 with respect to theslide is determined by the angular displacement = t.

Produce:1. the components of the absolute linear velocity of the particle 2 along the

system of coordinates XY Z and x1y1z12. the components of the absolute linear acceleration of the particle 2 along

the system of coordinates XY Z and x1y1z1


Solution

y1

x1

Y

X

z1 Z

i1

O

O

= t

= t

1 2

1 2

R

x1

j1 J

I

K=k1

R

Figure 11

According to Fig. 11, the absolute position vector of the particle is

R = i1R cost+ k1R sint (1.49)

The absolute velocity of the particle is the time derivative of this vector. To dier-entiate it one has to produce components of this vector along the inertial system ofcoordinates.

RX = I R = I i1R cost+ I k1R sint = R cost costRY = J R = J i1R cost+ J k1R sint = R sint cost (1.50)RZ = K R = K i1R cost+K k1R sint = R sint

Hence the components of the absolute velocity along the inertial system of coordinatesare

vX = R sint costR cost sintvY = R cost costR sint sint (1.51)vZ = R cost


The second derivative yields the components of the absolute acceleration along theinertial system of coordinates XY Z

aX = R2 cost cost+ 2R sint sintR2 cost costaY = R2 sint cost 2R cost sintR2 sint cost (1.52)aZ = R2 sint

Now, the components of the absolute velocity and the components of the absoluteacceleration can be transfer from the inertial system of coordinates XY Z to thex1y1z1systemof coordinates.

vx1 = i1 v == i1 I (R sint costR cost sint)

+i1 J (R cost costR sint sint)+i1 K (R cost)

= cost (R sint costR cost sint)+ sint (R cost costR sint sint)+0 (R cost)

= R sintvy1 = j1 v =R costvz1 = k1 v =R cost (1.53)

Similarly

ax1 = i1 a = R2 costR2 costay1 = j1 a = 2R sintaz1 = k1 a = R2 sint (1.54)


Problem 4

Y

X

1 2 P

O1

O2

1

2

1

2

Figure 12

Figure 12 shows the sketch of a Ferris wheel. Its base 1, of radius 1, rotatesabout the horizontal axis Z of the inertial system of coordinates XY Z. Its instanta-neous position is determined by the angular displacement 1. The wheel 2, of radius2, rotates about axis parallel to Z through the point O2. Its relative rotation isdetermined by the angular position 2.

Produce1. the equation of trajectory of the point PAnswer:X = 1 cos1 + 2 cos(1 + 2)Y = 1 sin1 + 2 sin(1 + 2)

2. the expression for the components of the absolute velocity of the point O2Answer:vO2X = 11 sin1vO2Y = +11 cos13. the expression for the components of the absolute acceleration of the point O2Answer:aO2X = 11 sin1 121 cos1aO2Y = +11 cos1 121 sin14. the expression for the components of the absolute velocity of the point PAnswer:vPX = 11 sin1 2(1 + 2) sin(1 + 2)vPY = +11 cos1 + 2(1 + 2) cos(1 + 2)5. the expression for the components of the absolute acceleration of the point P


Answer:aPX = 11 sin1121 cos12(1+ 2) sin(1+2)2(1+ 2)2 cos(1+2)aPY = +11 cos1121 sin1+2(1+ 2) cos(1+2)2(1+ 2)2 sin(1+2)


Problem 5

1

Y

2 l

3

X

Figure 13

The particle 3 and the rope 2 of length l (see Fig. 37) forms a pendulumoperating in the vertical plane XY of an inertial space. It is suspended from thecylinders 1 of radius The instantaneous position of the pendulum can be determinedby the angular displacement .

Produce1. the expression for the components of the absolute velocity of the particle 3 alongthe inertial system of coordinates XY Z as a function of the angular displacement Answer:vX = l sin+ ( sinvY = +l cos ( cos2. the expression for the components of the absolute acceleration of the particle 3along the inertial system of coordinates XY Z as a function of the angular displace-ment Answer:aX = l sin l2 cos+ (2 sin+ ( sin+ (2 cosaY = +l cos l2 sin (2 cos ( cos+ (2 sin

KINETICS OF PARTICLE 23

1.2 KINETICS OF PARTICLE

1.2.1 Equations of motionNewton second law for a single particle can be formulated in the following form

F =ddt(mr) (1.55)

whereF - is the resultant force acting on the particlem - is mass of the particler - is the absolute velocity.

If the mass is constant, the second law could be simplify to the following form

F = mr (1.56)

where r is called acceleration of the particle.If the position vector r and force F are determined by their components along

the same system of coordinates (eg r = IrX + JrY +KrZ , F = IFX + JFY +KFZ)the above equation is equivalent to three scalar equations.

FX = mrXFY = mrYFZ = mrZ (1.57)

If number of particles n is relatively low, we are able to produce a Free Body Diagramfor each of these particle separately and create 3n dierential equations that permiteach dynamic problem to be solved.1.2.2 Work

X

Y

Z

O

K

IJ

r(t)r(t+dt)r

rZ(t)

rY(t) rX (t)

d

F

Figure 14

Let us assume that the point of application of a force F slides along the path deter-mined by the position vector r (see Fig.14).


DEFINITION: It is said that the work done by the force F on the dis-placement dr is equal to the dot product of these two vectors.

dW = Fdr (1.58)If both, the force and the position vector are given by their components along theinertial system of coordinates XY Z

F = IFX(t) + JFY (t) +KFZ(t) r = IrX(t) + JrY (t) +KrZ(t) (1.59)

since dr isdr = vdt = (IrX(t) + JrY (t) +KrZ(t)) dt (1.60)

the work, according to the above definition, is

dW = (IFX(t) + JFY (t) +KFZ(t)) (IrX(t) + JrY (t) +KrZ(t)) dt= FX(t)rX(t) + FY (t)rY (t) + FZ(t)rZ(t)dt (1.61)

Hence, the work done between the instant t1 and t2 is

W =Z t2t1

(FX(t)rX(t) + FY (t)rY (t) + FZ(t)rZ(t)) dt (1.62)

If the force F is given as a function of the displacement s (see Fig. 15), we have

dW = Fdr =F cosds = Ftds (1.63)

X

Y

Z

O

s

t=0 tt v1

r(s)

F(s)

ds F(s)t

Figure 15

Hence, the total work done by the force between the position s1 and s2 is

W =Z s2s1

Ftds (1.64)


1.2.3 Kinetic energyDEFINITION: The scalar magnitude determined by the following expres-sion

T =1

2mv2 (1.65)

where m is mass of a particle and v is its absolute velocity, is called thekinetic energy of the particle.

X

Y

Z

O

K

IJ

r(t)rZ(t)

rY(t) rX (t)

m

v

Figure 16

If motion of the particle is given by means of the absolute position vector r(see Fig. 16) its velocity is

v = IrX(t) + JrY (t) +KrZ(t) (1.66)

The kinetic energy is

T =1

2mv2 =

1

2m(rX(t))

2 + (rY (t))2 + (rZ(t))

2 (1.67)


Principle of work and energy

X

Y

Z

O

s

t mt v1

r

F(s)

ds Ft

1a t

t2

Figure 17

Let F (see Fig. 17) be the resultant of all forces acting on the particle of mass m.According to the formula 1.63 we have

dW = Ftds (1.68)

Utilization of the second Newtons law yields the following expression for the work

dW = matds (1.69)

But the absolute tagential acceleration is

at =dvdt=dvdtdsds=dvdtvdtds

=vdvds

(1.70)

Introduction of expression 1.70 into 1.69 results in the following formula for the workdW

dW = mvdv (1.71)

Integration of the above equation within the range of time between the instant t1 andt2 yields

W =Z v2v1

mvdv =1

2mv22

1

2mv21 = T2 T1 (1.72)

The last relationship is known as the work and energy principle.

STATEMENT: The increment in the kinetic energy of a particle is equalto the work done by the resultant of all forces acting on the particle.


1.2.4 PowerDEFINITION: The rate of change of the work performed by a force iscalled power generated by this force.

P =dWdt

(1.73)

Since according to equation 1.58

dW = F dr (1.74)

the power is

P =F drdt

= F v (1.75)



X

Y

Ov

m

A

B

loo

vo

Figure 18

An element that can be considered as a particle of mass m (see Fig. 18) issuspended on the massless rope AOB in the vertical plane of the inertial space XY Z.The end B of this rope moves with the constant velocity v. When the length AO ofthe rope was equal to lo and its angular position was equal to o, the particle was freedto move with the initial velocity v0 = v. Produce the dierential equation of motionof the particle. Solve numerically the equations of motion and produce trajectory ofmotion of the particle for the following set of the numerical data. m = 1kg, lo = 1m,o = 1rad, v = 0.1m/s


Solution

X

Y

O v

m

r

l - vto

Figure 19

The instantaneous position of the particle is determined by the absolute po-sition vector r. Its components, according to Fig. 19, along the inertial system ofcoordinates are

r = I((l0 vt) sin) + J((l0 vt) cos) (1.76)Dierentiation of the position vector with respect to time yields vector of the absolutevelocity and the absolute acceleration of the particle.

r = I(l0 cos+ v sin+ vt cos) + J(l0 sin v cos+ vt sin) (1.77)

andr = I(aX) + J(aY ) (1.78)

where

aX = l0 cos+ l02 sin+ 2v cos+ vt cos vt2 sinaY = l0 sin l02 cos+ 2v sin+ vt sin+ vt2 cos (1.79)

Fig. 20 presents the free body diagram for the particle considered. Application ofthe Newtons law to this particle results in two equations with the two unknowns Rand .

maX = R sin

maY = mg R cos (1.80)

The first equation allows the unknown interaction force R to be determined.

R =maXsin

(1.81)

Introduction of the above expression into the second equation of 1.80 yields

maY = mg maXsin

cos (1.82)


X

Y

O

m

mg

R

Figure 20

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

X [m]

Y [m]

Figure 21


Introduction of the expressions 1.79 into equation 1.82 yields the dierential equationof motion that after simplification takes the following form

(lo vt) 2v+ g sin = 0 (1.83)

This equation must fulfil the following initial conditions

|t=0 = 0 and |t=0 = 0 (1.84)

Solution of the above equation as a function of time is presented in Fig. 21.The dotsin this diagram represent subsequent positions of the particle after equal intervals oftime equal to 0.02s.


Problem 7

2

1

Y

X

a

v

vt

O

Figure 22

The slide 1 ( see Fig. 22) rotates about the horizontal axis Z of the inertialsystem of coordinates XY Z. Its instantaneous position with respect to the inertialsystem of coordinates is determined by the angle (t). The slider 2 that can betreated as a particle of mass m moves along the slide with the constant velocity v.

Produce1. the expression for the absolute velocity of the particle 22. the expression for the absolute acceleration of the particle 23. the expression for the interaction force between the particle and the slide4. the expression for the driving force that must be applied to the particle to

move it with the assumed velocity v.5. the expression for the work produced by the gravity force


Solution

2

1

Y

X

a

v

vt

O

r

Figure 23

The absolute position vector r, according to the drawing in Fig. 23, is

r = I(a sin+ vt cos) + J(a cos+ vt sin) (1.85)

The vector of the absolute velocity can be obtained by dierentiation of the absoluteposition vector with respect to time.

v = I(a cos+ v cos vt sin) + J(a sin+ v sin+ vt cos) (1.86)

Similarly, the absolute acceleration is the time derivative of the absolute velocity.Hence

a = IaX + JaY (1.87)

where

aX = a cos+ a2 sin 2v sin vt sin vt2 cosaY = a sin a2 cos+ 2v cos+ vt cos vt2 sin (1.88)

The interaction force N and the driving force D (see the Free Body Diagram in Fig.24) can be obtained by application of the Newtons equation to the particle.


2

1

Y

X

a

vt

O

mg

N

D

Figure 24

Hence

maX = D cosN sinmaY = D sin+N cosmg (1.89)

Solving these equations with respect to the unknown driving force D and the normalinteraction force N one can get

D = m(aX cos+ aY sin+ g sin) (1.90)

andN = m(aX sin aY cos g cos) (1.91)

The work produced by the gravity force is

W =Z t0

J(mg) vd

=

Z t0

J(mg) I(a cos+ v cos v sin) ++J(a sin+ v sin+ v cos))d

=

Z t0

(mg)(a sin+ v sin+ v cos))d =

= mg(a cos+ v sin)|t0 (1.92)If (0) = 0

W = +mg (a(1 cos) vt sin) (1.93)This work, taken with sign -, is equal to the increment of the potential energy

of the particle.


Problem 8

2

1Y

X tO

Figure 25

The slide 1 (see Fig. 25) rotates with the constant angular velocity withrespect to the vertical plane XY of the inertial system of coordinates XY Z. Thebody 2 that can be treated as a particle of massm can move along the slide 1 withoutfriction. Produce the equations of motion of the particle.


Solution

2

1Y

X tO

l

X

Yr

Figure 26

Let l be the instantaneous distance between the particle and the origin of theinertial system of coordinates XY . The components of the absolute position vectorof the particle is

r = IX + JY = Il cost+ Jl sint (1.94)

Since l is a function of time its first and second derivative is

r = I(l cost l sint) + J(l sint+ l cost)r = I(l cost 2l sint l2 cost) + J(l sint+ 2l cost l2 sint)

(1.95)

2

1Y

X tO

l

X

Y

mg

R

Figure 27

The second derivative represents the absolute acceleration of the particle.Hence, the application of the Newtons equation to the free body diagram shownin Fig. 27 yields.

m(l cost 2l sint l2 cost) = R sintm(l sint+ 2l cost l2 sint) = R costmg (1.96)

Elimination of the unknown reaction R from the above set of equations allows theequation of motion to be produced in the following form.

l l2 = g sint (1.97)


Since the equitation is linear with respect to the distance l, it can be analyticallysolved. Its solution is

l = Aet +Bet +g22

sint (1.98)

where A and B are constant magnitudes. For the following initial conditions

l | t=0 = 0l | t=0 = 0 (1.99)

the constants A and B are

A = g42

; B =g42

(1.100)

and the equation of motion takes form

l = g42

et +g42

et +g22

sint (1.101)

The parametric equations of the trajectory of the particle are

X = g42

et +g42

et +g22

sintcost

Y = g42

et +g42

et +g22

sintsint (1.102)

It is presented in Fig. 28 for = 1

-50

0

50

100

150

200

250

300

350

-80 -60 -40 -20 0 20 40 60 80 100X

Y

Figure 28


Problem 9

H

1

2

L

w

3

A

B

Figure 29

The parabolic slide 1 (see Fig. 29) is motionless with respect to the verticalplane of the inertial space. The pin 3 of mass m follows this parabolic slide and it isdriven by the slide 2. The slide 2 moves with the constant velocity w. The frictionbetween the slides and the pin 3 can be neglected. Produce

1. the expression for the interaction force between the pin 3, and the bothslides 1 and 2.

2. the kinetic energy of the particle at the position A and B.3. the work produced by all the forces acting on the particle4. verify your results by means of the work-energy principle


Solution

H

1

2

L

3

A

B

G

N

R

X

Y

Figure 30

Let us take advantage of the second Newtons law

ma = F (1.103)

The absolute acceleration a according to the equation 1.46 is

a = I(0) + J(2bw2) (1.104)

The resultant F of all forces acting on the particle, according to the Free BodyDiagram shown in Fig. 30 is

F = G+N+R (1.105)

whereG = J(mg) (1.106)N = n1N (1.107)

R = IR (1.108)

According to 1.45

n1 =1

1 + 4b2w2t2(I (2bwt) + J (1)) (1.109)

Hence the equation 2.47 can be rewritten in the following form

J2bw2m = J (mg) + N1 + 4b2w2t2

(I (2bwt) + J (1)) + IR (1.110)


This vector equation is equivalent to two scalar equations

0 =2bwtN1 + 4b2w2t2

+R

2bw2m = mg + N1 + 4b2w2t2

(1.111)

Solving these equations with respect to the unknown N and R we have

N =1 + 4b2w2t2m(2bw2 + g)

R = 2bwm(2bw2 + g)t (1.112)

The kinetic energy of the particle is

T =1

2m(v(t))2 (1.113)

The time associated with the position A is tA = 0. Hence the kinetic energy at theposition A according to 1.40 is

TA =1

2mwq1 + 4b2w2t2A

2=1

2mw2 (1.114)

The time associated with the position B is tB = Lw . Hence the kinetic energy at theposition B is

TB =1

2mwq1 + 4b2w2t2B

2=1

2mw2

1 + 4b2L2

(1.115)

Since the normal force N is perpendicular to the trajectory of the particle, the workdone by this force is equal to zero.

WN = 0 (1.116)

The work produced by the force R is

WR =Z tBtA

Rdr =Z tBtA

R vdt (1.117)

According to 1.39

v = r = Iw + J2bw2t (1.118)

Hence

WR =Z tBtA

I2bwm(2bw2 + g)t Iw + J2bw2t dt = Z tBtA

2bw2m(2bw2 + g)tdt

= bmL2(2bw2 + g) (1.119)


Similarly

WG =Z tBtA

Gdr =Z tBtA

J (mg) Iw + J2bw2t dt = Z tBtA

2mgbw2t

dt = mgbL2

(1.120)The work-kinetic energy principle says that

WN +WR +WG = TB TA (1.121)

Indeed

0 + bmL2(2bw2 + g) +mgbL2

=1

2mw2

1 + 4b2L2

12mw2 (1.122)


Problem 10

v

m

r

Figure 31

The axis of symmetry the cylinder shown in Fig. 31 coincides the horizontalaxis of an inertial system of coordinates. Its radius is equal to r. The initial velocityof the ball of mass m at the position shown in Fig. 31 is v. It rolls over the innersurface of the cylinder. Produce the equation of motion of the ball and the expressionfor the normal interaction force between the ball and the cylinder.


Solution

v

m

rX

Y

r

N

G

Figure 32

Motion of the ball is determined by the position vector r.(see Fig. 32)

r = Ir cos+ Jr sin (1.123)

where is an unknown function of time. Two subsequent dierentiations yields theabsolute acceleration of the ball.

r = I (r sin) + J (r cos) (1.124)

r = Ir sin r2 cos

+ J

r cos r2 sin

(1.125)

Application of the second Newtons law to the ball yields

mr = G+N (1.126)

whereG = J(mg) N = I(N cos) + J(N sin) (1.127)

Introducing the expressions 1.125 and 1.127 into the equation 1.126one may get

Ir sin r2 cos

m+ J

r cos r2 sin

m

= J(mg) + I(N cos) + J(N sin) (1.128)

This equation is equivalent to the two following scalar equationsr sin r2 cos

m = N cos

r cos r2 sinm = N sinmg (1.129)

The second equation of the set 1.129 allows the normal force to be obtained

N = mgsin

cossin

mr+mr2 (1.130)


Upon introduction of the above expression into the first equation of 1.129, one canget the following equation of motion

+grcos = 0 (1.131)

The initial conditions are as follows

|t=0 = 2

|t=0 = vr (1.132)

Solution of the equation of motion 1.131 with the initial conditions 1.132 if substitutedinto expression 1.130 yields expression for the normal force N . As long as the normalforce is positive, the ball is in contact with the cylinder. N = 0 flags the instant oftime at which the ball is loosing contact with the cylinder.


Problem 11

Yy

X

x

r

rx

ry

A

z

Z

Ao

2

1

Figure 33

The relative motion of the shuttle 1 with respect to the batten 2 of a loom(Fig. 33) is given in the rotating system of coordinates x, y, z by the position vectorr.

r = irx + jry (1.133)

where:rx = A sin, ry = Ao A cos (1.134)

The rotation of the batten 2 about axis X is determined by its angular dis-placement .

Produce1. the expressions for components of the absolute velocity and acceleration of

the shuttle 1 along the absolute (XY Z) and intrinsic ( tnb) system of coordinates2. the equation of motion of the shuttle


Solution.According to Eq. 1.133 and Eq. 1.134 the position vector r is

r = irx + jry = iA sin + j(Ao A cos) (1.135)

The components of the absolute velocity and acceleration may be obtained by dif-ferentiation of components of the absolute position vector r along inertial system ofcoordinates XY Z. These components may be obtained as the dot products of thevector r and the corresponding unit vectors IJK.

rX = I r = I (irx + jry) = rx = A sinrY = J r = J (irx + jry) = cos ry = (Ao A cos) cosrZ = K r = K (irx + jry) = cos(90o )ry = (Ao A cos) sin(1.136)

Dierentiation of the above components with respect to time yields the componentsof the absolute velocity.

rX = A cos

rY = (A sin) cos (Ao A cos ) sinrZ = (A sin) sin+ (Ao A cos ) cos (1.137)

Similarly, dierentiation of the components of the absolute velocity allows the com-ponents of the absolute acceleration to be obtained

rX = A cos A2sin

rY = (A sin +A2cos) cos (Asin) sin (A sin) sin

(Ao A cos) sin (Ao A cos)2 cosrZ = (A sin +A

2cos) sin+ (A sin) cos+ (A sin) cos

+(Ao A cos) cos (Ao A cos)2 sin (1.138)

Components of the absolute acceleration along system of coordinates xyz

r = iax + jay + kaz (1.139)

may be obtain as a dot product of r and unit vectors i, j,k

ax = r i = I irX + J irY +K irZ = rX= A cos A2 sin

ay = r j = I jrX + J jrY +K jrZ = rY cos+ rZ sin= A sin +A

2cos (Ao A cos)2

az = r k = I krX + J krY +K krZ = rY sin+ rZ cos= 2A sin + (Ao A cos) (1.140)

To employ Newtons equationmr =

XF (1.141)


let us introduce the intrinsic system of coordinates t,n,b shown in Fig. 34.

Yy

X

x

z

Z

t

n

Fn

G

Fb

Ftb

n

bt o

Figure 34

The unit vector t coincides with the relative velocity of the shuttle r0. Hence

t =r0

r0(1.142)

wherer0 = irx + jry = iA cos + jA sin (1.143)

r0 =qA2

2cos2 +A2

2sin2 = A (1.144)

Introducing Eqs.. 1.143 and 1.144 into Eq. 1.142 one may gain

t = i cos + j sin (1.145)

The binormal unit vector b is perpendicular to the plane xy hence

b = k (1.146)

The normal unit vector n may be calculated as a cross product of b and t.

n = b t = i j k0 0 1cos sin 0

= i sin + j cos (1.147)


Hence, the components of acceleration along the intrinsic system are

at = t r = (i cos + j sin) (iax + jay + kaz) = ax cos + ay sin= A (Ao A cos)2 sin

an = n r = (i sin + j cos) (iax + jay + kaz) = ax sin + ay cos = A

2 (Ao A cos)2 cos ab = b r = k (iax + jay + kaz) = az

= 2A sin + (Ao A cos ) (1.148)Now, the Newtons equations

ma = F

may be written in the following form

mab = Fb +Gbmat = Ft +Gtman = Fn +Gn (1.149)

where Gb, Gt and Gn are components of the gravity force along the intrinsic systemof coordinates. They are

Gb = Jmg b =mgJ k = mg cos(J,k) = +mg sinGt = Jmg t =mgJ (i cos + j sin) = mg sin cos (1.150)Gn = Jmg n =mgJ (i sin + j cos) = mg cos cos

There is a relationship between the friction force Ft and the interaction forces Fb andFn. Namely

Ft = bFb + nFn (1.151)

where b and n are the friction coecients between the walls and the shuttle.Introduction of 1.151 into 1.149 yields

mab = Fb +Gbmat = bFb + nFn +Gtman = Fn +Gn (1.152)

The above equations are linear with respect to the interaction forces Fb and Fn.Thereforethey can be easily eliminated from the equations 1.152.

mat = b (mab Gb) + n (man Gn) +Gt (1.153)Introduction of 1.148 and 1.150 into 1.153 yields the dierential equation that governmotion of the shuttle.

mA (Ao A cos)2 sin

= b(m

2A sin + (Ao A cos)

mg sin)

+n(mA

2 (Ao A cos)2 cos+mg cos cos)mg sin cos

Its solution (t) approximate motion of the shuttle.


Problem 12

Y

screw line

X

r

Z

rYrZ rX

Oa

s

H

F

Figure 35

The bead of mass m, which can be considered as a particle, moves along thescrew line shown in Fig. 35. Its motion with respect to the inertial space XY Z isgiven by the following position vector

r = Ia cost+ Ja sint+Kbt (1.154)

Produce expression for1. the interaction force between the spiral slide and the bead.Answer:R = I(ma2 cost+ mgaba22+b2 sint)+J(ma2 sint+

mgaba22+b2 cost)+K(

mga22

a22+b2 )2. the force F which must be applied to the bead to ensure its motion according tothe equation 1.154Answer:F = I( mgaba22+b2 sint) + J(

mgaba22+b2 cost) +K(

mgb2

a22+b2 )

3. the work produced by individual forces acting on the bead for 0 < t < 2Answer:WR = 0 WF =

2mgb WG =

2mgb

4. the kinetic energy of the particle as a function of timeAnswer:T = 1

2m(a22 + b2)


Problem 13

Y

X

1 2 P

O1

O2

1

2

1

2

Figure 36

Figure 36 shows the sketch of a Ferris wheel. Its base 1, of radius 1, rotatesabout the horizontal axis Z of the inertial system of coordinates XY Z. Its instanta-neous position is determined by the angular displacement 1. The wheel 2, of radius2, rotates about axis parallel to Z through the point O2. Its relative rotation isdetermined by the angular position 2. A person that can be treated as a particle ofthe mass m is seating at the wheel 2. Motion of the Ferris wheel is restricted to thevertical plane XY.

Produce1. the expression for the work produced by the gravity force acting on the particle PAnswer:W = mg(1 sin1 + 2 sin(1 + 2))2. the expression for the kinetic energy of the particle PAnswer:T = 1

2m21

21 +

22

21 +

22

22 + 2

2212 + 21212 cos2 + 212

21 cos2

3. the expression for the components of the interaction force between the particle Pand the wheelAnswer:RX = maPXRY = maPY +mgwhereaPX = 11 sin1121 cos12(1+ 2) sin(1+2)2(1+ 2)2 cos(1+2)aPY = +11 cos1121 sin1+2(1+ 2) cos(1+2)2(1+ 2)2 sin(1+2)


Problem 14

1

Y

2 l

3

X

Figure 37

The particle 3 is suspended from the cylinders 1 of radius by means of themassless rope 2 of length l (see Fig. 37). Its instantaneous position can be determinedby the angular displacement .Motion of the particle is restricted to the vertical planeXY of an inertial system of coordinates.

Produce1. the dierential equation of motion of the particle 3Answer:aX aY cot g = 0whereaX = l sin l2 cos+ (2 sin+ ( sin+ (2 cosaY = +l cos l2 sin (2 cos ( cos+ (2 sin2. the expression for the tension in the rope as a function of the angular position Answer:T = maY 1sin


Problem 15

y

x Y

X

t

1 23

v L

O

Figure 38

The slider 2, shown in Fig. 38, can be moved inward by means of the string3 while the slotted arm 1 rotates about the horizontal axis Z of the inertial systemof coordinates XY Z. The relative velocity v of the slider with respect to the arm isconstant and the absolute angular velocity of the arm is constant too. The slider2 can be considered as a particle of mass m. When the arm coincides with the axisX the slider is L distant from the origin O. The system of coordinates xyz is rigidlyattached to the arm 1.

Produce:1. the expression for the components of the absolute linear velocity of the slider 2along the system of coordinates xyzAnswer:

vs = i( v) + j(L vt)2. the expression for the components of the absolute linear acceleration of the slider2 along the system of coordinates xyzAnswer:as = i( 2(L vt)) + j(2v)3. the expression for the components of the absolute linear acceleration of the slider2 along the system of coordinates XY ZAnswer:as = I( 2(L vt) cost+ 2v sint) + J( 2(L vt) sint 2v cost)4. the expression for the interaction force between the slider 2 and the arm 1


Answer:N = mg cost 2mv5. the expression for the tension in the string 3Answer:T = m2(L vt)mg sint


Problem 16

Y

X

t1

2

1

X

Y

Figure 39

A skier that can be treated as the particle 1 of mass m (see Fig. 39) is skiingdown the slope 2. The shape of slope is determined by the following equation

Y = aX2

There is a dry friction between the skier 1 and the slope 2. The friction coecient is.

Produce:1. the expression for the components of the tangential unit vector t1Answer:t1 = I

11+4a2X2

+ J 2aX1+4a2X2

2. the expression for the gradient .Answer: = arccos 1

1+4a2X2

3. the free body diagram for the skier 1.4. the dierential equations of motion of the skierAnswer:2maX2 + 2amXX = mg + mX(cos+ sin) cossinwhere cos = 1

1+4a2X2sin = 2aX

1+4a2X2


Problem 17

L = vt

= t

X 1

Y 23

a

y

x

Figure 40

The belt 2 of the belt conveyor 1 shown in Fig. 40 moves with a constantvelocity v. Simultaneously, its boom is being raised with a constant angular velocity. The object 3 that can be considered as a particle of mass m is motionless withrespect to the belt 2.

Produce:1. The expression for the absolute linear velocity of the object 32. The expression for the tangential, normal and binormal unit vectors associatedwith the absolute trajectory of motion of the object 33. The expression for the absolute linear acceleration of the object 34. The expression for the interaction forces between the object 3 and the belt 2


Solution1.

L = vt

= t

X

1

Y 23

a

y

x

L

a

r

Figure 41

The absolute position vector of the object 3 (see Fig. 41) is

r = a+ L = j(a) + i(vt) (1.155)

It components along the inertial system of coordinates XY Z are

rX = I r = I (j(a) + i(vt)) = a sint+ vt costrY = J r = J (j(a) + i(vt)) = a cost+ vt sint (1.156)

The components of the absolute velocity v of the object are

vX = rX = a cost+ v cost vt sintvY = rY = a sint+ v sint+ vt cost (1.157)

Magnitude of the absolute velocity is

v =qv2X + v

2Y =

q(v a)2 + (vt)2 (1.158)

2.The tangential unit vector associated with the absolute trajectory is

t1 =vv =

1(va)2+(vt)2

I(a cost+ v cost vt sint)+ J(a sint+ v sint+ vt cost) =

= I (t1X) + J (t1Y )

(1.159)

Since the absolute trajectory is placed in the the plane XY , the binormal unit vectoris perpendicular to this plane

b1 = K (1) (1.160)


Hence, the normal unit vector is

t1 n1 = b1(t1 n1) t1 = b1 t1t1 (t1 n1) = b1 t1

t1 (n1 t1) + n1 (t1 t1) = b1 t1n1 = b1 t1

n1 = b1 t1 = I J K0 0 1t1X t1Y 0

=

= I (t1Y ) + J (t1X) (1.161)

3.The absolute acceleration as the fist derivative of the absolute velocity vector

isa = IaX + Ja (1.162)

where

aX = vX = 2v sint vt2 cost+ a2 sintaY = vY = 2v cost vt2 sint a2 cost (1.163)

4.

L = vt

= t

X 1

Y 23

a

y

x

G

F

N

Figure 42

Application of the Newtons law to the FBD shown in Fig. 42 yields

maX = N sint+ F costmaY = N cost+ F sintG (1.164)


Hence, the interaction forces are

N = maX sint+maY cost+G costF = maX cost+maY sint+G sint (1.165)

Chapter 2

PLANE DYNAMICS OF A RIGID BODY

2.1 PLANE KINEMATICS OF A RIGID BODY

DEFINITION: A body which by assumption is not deformable and there-fore the distances between its points remains unchanged, regardless offorces acting on the body, is called rigid body.

DEFINITION: If there exists such a plane XY of an inertial space XY Zthat trajectories of all points of the body considered are parallel to thisplane, motion of the body is called plane motion and planeXY is called planeof motion.

Y

XO

o x

Z

y

z z

y

x

body system of coordinatesinertial system of coordinates

plane of motion

Figure 1

In this chapter we shall assume that such a plane of motion exists.To analyze motion of a rigid body we attach to the body a system of coordi-

nates xyz at an arbitrarily chosen point o (see Fig.1). Such a system of coordinatesis called body system of coordinates. Axis z of the body system of coordinates ischosen to be parallel to the axis Z of the inertial system of coordinates as is shown

PLANE KINEMATICS OF A RIGID BODY 60

in Fig 1. Hence, the plane xy of the body system of coordinates is always parallel tothe plane XY . Furthermore, without loosing the generality of consideration, we canassume that the origin of the body system of coordinates and origin of the inertialsystem of coordinates are in the same plane. Hence, motion of the rigid body can beuniquely determined by motion of its body system of coordinates. The body systemof coordinates, in a general case, can not be classified as the inertial one

2.1.1 Non-inertial (moving) systems of coordinates

DEFINITION: System of coordinates which can not be classified as iner-tial is called non-inertial system of coordinates.

The non-inertial systems of coordinates are denoted by lower characters (e.g. xy) todistinguish them from inertial ones which are usually denoted by upper characters(e.g. XY ).

DEFINITION: If a non-inertial system of coordinates does not rotate (itsaxes xy are always parallel to an inertial system) the system is called translatingsystem of coordinates (Fig. 2 a).

DEFINITION: If the origin of a non-inertial system of coordinates coin-cides always the origin of an inertial system of coordinates, the non-inertialsystem is called rotating system of coordinates (Fig. 2b)

In a general case a non-inertial system of coordinates may translate and rotate.

DEFINITION: System of coordinates which can translate and rotate iscalled translating and rotating system of coordinates (Fig. 2c)

Y

XO o

x

y

X

Y

Oo

x

y Y

XO

o x

y

a b c

Figure 2


Translating system of coordinates

X

Y

O o x

y

ro

Figure 3

Since axes of the translating system of coordinates are always parallel to the axesof the inertial system of coordinates, the components of a vector along the non-inertial system of coordinates are equal to the components of the inertial systemof coordinates. To determine motion of the translating system of coordinates it issucient to introduce one position vector ro (Fig. 3).

Rotating system of coordinates

Y

XO o

x

y

j i

I

J

Figure 4

Since axes of the rotating system of coordinates change their angular position withrespect to the inertial system of coordinates, the components of a vector along theinertial and the rotating system of coordinates are not equal to each other. Thereforeit is essential to develop a procedure that allows the components of a vector along theinertial system of coordinates to be produced if the vector is given by its componentsalong the rotating system of coordinates.

Matrix of direction cosines


Y

X

Oo

x

y

j i

I

J

P

rrx ry

rX

rY

Figure 5

Let us assume that the vector r (Fig. 5) is given by its components along therotating system of coordinates xyz.

r = irx + jry (2.1)

Components of the vector r along the inertial system of coordinates XY may beobtained as the scalar products of the vector r and the unit vectors IJ.

rX = r I = rxi I+ ryj I = rx cosiI+ ry cosjIrY = r J = rxi J+ ryj J = rx cosiJ+ ry cosjJ (2.2)

The last relationship can be written in the following matrix formrXrY

=

cosiI cosjIcosiJ cosjJ

rxry

=

cos sinsin cos

rxry

= [Cri]

rxry

(2.3)

The transfer matrix [Cri] is called matrix of direction cosines.If we assume that the vector r is given by its components along the inertial

system of coordinates and repeat the above derivation, we are gettingrxry

=

cos sin sin cos

rXrY

= [Cir]

rXrY

(2.4)

Hence one can conclude that the transfer matrix of the direction cosines is equal tothe inverse matrix of the direction cosines

[Cri]1 = [Cri]T = [Cir] (2.5)

Another useful relationship should be noticed from Fig. 6. Cosine of angle betweentwo unit vectors e.g. i and J is equal to the component of one on the other.

iJJ

iJ

iY

x

Figure 6


cosiJ = iYi= iY =

JxJ= Jx (2.6)

X

Y

O

iiX

iY

Figure 7

From Fig. 7 one can see that components iX,iY ,which are equal to corre-sponding direction cosines, fulfil the following relationship.

i2X + i2Y = cos

2iI+ cos2iJ = 1 (2.7)

Angular velocity and angular acceleration

Y

X

O o

x

y

j

i

I

J

Z,z

K,k

y

x

d

d

Figure 8

It is easy to show that the infinitesimal angular displacement d can be con-sidered as a vectorial magnitude. The vector of the infinitesimal angular displacement


is perpendicular to the plane of rotation ( XY ) and its sense is determined by theright-hand rule (see Fig. 8). Such vectors which have determined only direction andsense only, are called free vectors to distinguish them from linear vectors (sense andline of action is determined) and position vectors ( position of its tail, direction andsense is determined). The angular displacement is considered positive if the sense ofits vectorial representation coincides with the positive direction of the axis z.

Y

X

O o

x

y

Z,z

y

x

d

d

d

r

rd

P

Figure 9

Now, let us consider point P that is fixed with respect the xyz system ofcoordinates. Its position is determined by the position vector r (Fig. 9). Let d bean infinitesimal angular displacement of the system of coordinates xyz. Hence, thepoint P at the instance considered moves along the circle of radius r. The infinitesimalincrement dr of vector r is tangential to the circle. Its scalar magnitude is

dr = r d (2.8)

Since1. vector dr is perpendicular to the plane formed by vectors d and r2. magnitude of the vector dr is r d sin 90o

3. vectors d, r, dr forms the right handed system of vectorsthe vector dr can be considered as the vector product of vector d and r.

dr = d r (2.9)The velocity of the point P is

v =drdt=ddt r (2.10)


DEFINITION: The vector ddt is called vector of the absolute angular velocity of the rotating system of coordinates xyz.

= ddt

(2.11)

According to 2.11, the vector of the angular velocity possesses all properties of thevector d.The velocity of P can be expressed as follow.

v = r (2.12)The angular acceleration is defined as the first derivative of the vector of the angularvelocity with respect to time.

= ddt= (2.13)

Derivative of a vector expressed by means of components along arotating system of coordinates

Motion of the rotating system of coordinates xy with respect to the inertialone can be defined by the angular displacement or, alternatively, the rotationalmotion can be determined by its initial position and the vector of its angular velocity.

Let be the absolute angular velocity of the rotating system of coordinatesxyz (see Fig. 10).

Y

X

O o

x

y

ji

I

J

Z,z

K,k

A

Ax

Ay

Figure 10


Consider a vectorA that is given by its components along the rotating systemof coordinates xy.

A = iAx + jAy (2.14)

Let us dierentiate this vector with respect to time.

A =ddt(iAx + jAy) (2.15)

HenceA = iAx + jAy + iAx + jAy (2.16)

The first two terms represent vector which can be obtained by direct dierentiatingof the components Ax, Ay with respect to time. This vector will be denoted by A0

A0 = iAx + jAy (2.17)

ThusA = A0 + iAx + jAy (2.18)

According to the definition of the vector derivative, the first derivative of the unitvector i is the ratio of the infinitesimal vector increment di and dt (Fig. 11).

i =didt=vidtdt

= vi (2.19)

were vi is a velocity of the head of the vector i (see Fig.11).

Y

XO o

x

y

j i

I

J

v

i

Figure 11

But, according to Eq. 2.12

vi = i (2.20)

Hencei = i (2.21)

Similarlyj = j (2.22)

Introducing the above expressions into Eq. 2.18 we have

A = A0 + iAx + jAy = A0 + (iAx + jAy) (2.23)


and eventually one may gainA = A0 + A (2.24)

where:A0 = iAx + jAy - is the absolute angular velocity of the rotating system of coordinates xy

along which the vector A was resolved to produce vector A0

A - is the dierentiated vector.The last formula provides the rule for dierentiation of a vector that is resolved alonga non-inertial system of coordinates. It can be applied to any vector (position vectors,linear vectors, free vectors).

Translating and rotating system of coordinates

Y

XO

o x

y

ro

Figure 12

Motion of the translating and rotating system of coordinates xyz shown in Fig. 12,can be uniquely determined by means of the two vectors ro and. The position vectorro determines motion of its origin with respect to the inertial system of coordinatesXY Z and the vector of the angular velocity determines its rotation.


2.1.2 Motion of rigid bodyRotation of rigid body about fixed axis

Y

X O o

x

y

j i

I

Z,z

K,k

J

Figure 13

DEFINITION: If there exists such axis of a rigid body that does not move

with respect to the inertial system of coordinates, it is said that the bodyperforms rotation about this axis and the axis is called axis of rotation.

Let us assume that the system of coordinates xyz shown in Fig. 13 rotates about axisZ of the inertial system of coordinates XY Z. Its rotational motion can be uniquelydetermined by the angular displacement or, alternatively, by the angular velocity. Let us attach to this system of coordinates a rigid body. According to the abovedefinition the body performs rotation about axis Z.

DEFINITION: The angular velocity of the system of coordinate xyz iscalled angular velocity of the rigid body that is attached to the system of coor-dinates.

The vector of the angular velocity of the rigid body is

= K = k (2.25)

The angular acceleration of the body is the time derivative of the vector of the angularvelocity. Since the vector is always perpendicular to the plane of motion of body,its derivative is.

= = K ddt = K (2.26)


Y

X Oo

x

y

Z,z

P r PA

rA

r P

A

Figure 14

Let us consider an arbitrarily chosen point P that belong to this body. Itsabsolute position vector rP has two components rA and rPA (see Fig. 14).

rP = rA + rPA (2.27)

The vector rA determines the plane of motion of the point P whereas the vector rPArepresents distance of the point P from the axis of rotation z.

rA = kz = Kz; rPA = ix+ jy (2.28)

Here, x, y and z are the coordinates of the point P with respect to the body systemof coordinates xyz.

The absolute velocity of the point P is the time derivative of the absoluteposition vector rP .

vP = rP = rA + rPA (2.29)

Since coordinates x, y and z are time independent,

rA = 0

rPA = r0PA + rPA = rPA (2.30)

Introduction of equations 2.30 into 2.29 yields

vP = rPA (2.31)


As one can see from the last relationship, the velocity of the point P does not dependon choice of the plane of motion of the point P . Therefore in the further development,without any harm to generality of the consideration, we will be assuming that theplane of motion coincides with the plane XY (see Fig.15.). In this case rP = rPA.Hence

vP = rP (2.32)

Y

X

O o

x

y

P

xy

rP

vP aPt

aPn

Figure 15

From the relationship 2.32 one can see that the vector of velocity must beperpendicular to both and rP as it is shown in Fig. 15. The absolute accelerationcan be obtained by dierentiation of the absolute velocity vector vP .

aP = vP =ddt( rP ) = rP + vP = rP + vP (2.33)

The first term as the cross-product of and rP must be perpendicular to both of themand it is called tangential component of acceleration aPt. The second one must beperpendicular to velocity and is called normal component of acceleration aPn. Theirmagnitudes are as follows

aPt = rP = rPaPn = vP = 2rP = 2rP (2.34)


General motion of rigid body

X

Y

O

y

x

o

A

rA

ro

rAo

B

rBA

rBorB

Figure 16

Motion of the translating and rotating system of coordinates xyz shown in Fig. 16can be uniquely determined by means of two vectors ro and . The position vectorro determines motion of its origin with respect to the inertial system of coordinatesXY Z and the vector of angular velocity determines its rotation.DEFINITION: It is said that a body rigidly attached to the translatingand rotating system of coordinates xyz performs the general motion and theangular velocity of the system of coordinates xyz is called angular velocityof the rigid body.

Angular acceleration of the body as the first time derivative of the vector is

= = ddt(K) = K (2.35)

The absolute velocity of the arbitrarily chosen point A is the first time derivative ofits absolute position vector rA.

vA = rA =ddt(ro + rAo) = ro + rAo (2.36)

If the vector rAo is resolved along the body system of coordinates xyz, its componentsare time independent and therefore its derivative is

rAo = rAo (2.37)Hence

vA = ro + rAo (2.38)


In the same manner one can prove that the absolute velocity of the point B is

vB = ro + rBo (2.39)The relative velocity vBA of the point B with respect to the point A is then

vBA = vB vA = rBo rAo = (rBo rAo) = rBA (2.40)Hence the relative velocity between two points that belong to the same rigid body isalways perpendicular to the line joining them and its scalar magnitude is

vBA = rBA sin 90o = rBA (2.41)

The sense of this vector, according to the definition of the cross product, can bedetermined by rotation of the vector rBA by 90o in the direction of the angularvelocity (see Fig. 17). It can be seen from the equation 2.40 that the absolutevelocity of an arbitrarily chosen point B is

vB = vA + vBA = vA + rBA (2.42)

The absolute acceleration of the point A one can get by dierentiation of thevector of absolute velocity

aA = vA = ro +ddt( rAo) = ro + rAo + ( rAo) (2.43)

Similarly, the absolute acceleration of the point B is

aB = vB = ro +ddt( rBo) = ro + rBo + ( rBo) (2.44)

The dierence between these two above vectors produces the relative accelerationaBA.

aBA = aB aA = rBo + ( rBo) rAo ( rAo) == (rBo rAo) + ( rBo rAo) == rBA + ( (rBo rAo)) = (2.45)= rBA + ( rBA) = rBA + vBA

The normal component of the relative acceleration

aBAn = vBA (2.46)and its tangential component

aBAt = rBA (2.47)are shown in Fig. 18.

Eventually we can state that if A and B belong to the same body the rela-tionship between their absolute accelerations is

aB = aA + aBA = aA + rBA + vBA (2.48)


X

Y

O

y

x

o

A

rA

ro

rAo

B

rBA

rBorB

vBA

Figure 17

X

Y

O

y

x

o

A

rA

ro

rAo

B

rBA

rBorB

vBA

aBAtaBAn

Figure 18



Y v

Z

C

a

c

l

2

1

3 A

B G

A

Figure 19

The bus 1 (see Fig. 19) moves in the vertical plane Y Zof the inertial systemof coordinates XY Z. The rear wheels 2 of the bus are driven with the constant linearvelocity vA The bus is climbing on the hump 3 which is inclined to the horizontalplane at the angle .

Develop the expression for1. the absolute angular velocity of the bus2. the absolute angular acceleration of the bus3. the absolutethe absolute velocity of the front axle B4. the absolute linear acceleration of the front axle B5. the absolutethe absolute velocity of the centre of gravity G6. the absolute linear acceleration of the centre of gravity G


Y

Z

C

a

c

l

2

1

3 A

B G

l

O

y

z

x

Figure 20

SolutionThe origin of the inertial system of coordinates Y Z has been chosen to coincide

with the axle A when the bus reaches the hump (point B coincides the point C - seeFig. 20). Hence, for any arbitrarily chosen instant of time, the distance OA is

AC = l vAt (2.49)

The system of coordinates xyz is rigidly attached to the bus. Hence the angle represents the absolute angular displacement of the body 1 system of coordinates.Both the angular displacement and the distance CB = x can be produced by solvingthe following vector equation

AB =AC +

CB (2.50)

whereAB = jlAC = J(l vAt) (2.51)CB = Jx cos+Kx sin

Hencejl = J(l vAt) + Jx cos+Kx sin (2.52)

This equation is equivalent to two scalar equations that can be obtained by subsequentmultiplication of the equation 2.52 by the unit vectors J and K.

l cos = l vAt+ x cosl sin = x sin (2.53)

This set has the following solutions

= arcsinl vAt

lsin

(2.54)


Y

Z

C

a

c

l

2

1

3 A

B G

l

O

y

z

x

Figure 21

and

x =l sinsin

(2.55)

Hence the absolute angular velocity is

= I (2.56)

and the absolute angular acceleration is

= I (2.57)

To get the absolute linear velocity of the point B ane has to dierentiate the absoluteposition vector rB (see Fig. 21)

rB = J(vAt+ l cos ) +Kl sin (2.58)

HencevB = J(vA l sin) +Kl cos (2.59)

The derivative of the absolute velocity yields the absolute acceleration

aB = J( l sin l2cos) +K(l cos l2 sin) (2.60)

Dierentiation of the absolute position vector of the point G

rG = J(vAt+ a cos c sin) +K(a sin + c cos) (2.61)

yields the absolute velocity of the centre of gravity G

vG = rG = J(vA a sin c cos) +K(a cos c sin) (2.62)

The absolute acceleration of the centre of gravity is

aG = vG = J( a sin a2cos c cos + c2 sin) + (2.63)

+K(a cos a2 sin c sin c2 cos)


Problem 19

Y

X C

2 1 B

(t)G

A

r

3

d a

b

Figure 22

Motion of the crankshaft 1 shown in Fig. 22 is given by its angular position(t). Produce the expression for the absolute velocity and the absolute accelerationof its piston 3.

Given are (t), r, a, b, d


SolutionThe instantaneous position of the system ( X and ) can be obtained by

solving the following vector equation (see Fig. 23).

Y

X C

2 1 B

(t)G

A

r

3

d a

b

X

Figure 23

AB +

BC = J(d) + IX (2.64)

or

Ir cos+ Jr sin+ I((a+ b) cos) + J((a+ b) sin) = J(d) + IX (2.65)This equation is equivalent to two scalar equations

r cos+ (a+ b) cos = X

r sin (a+ b) sin = d (2.66)Hence

sin =d+ r sina+ b

(2.67)

and

X = r cos+ (a+ b)

s1

d+ r sina+ b

2(2.68)

Velocity of the point B is the vector derivative of the position vectorAB.

vB =ddt(Ir cos+ Jr sin) = I (r sin) + J (r cos) (2.69)

According to the relationship between two points of the same body the linear velocityof the point C is

vC = vB + vCB = vB +2BC =

= I (r sin) + J (r cos) +

I J K0 0 2(a+ b) cos (a+ b) sin 0

=

= I (r sin+ (a+ b)2 sin) + J (r cos+ (a+ b)2 cos) (2.70)


According to the constraints shown in Fig. 22, the component of the velocity vCalong axis Y must be equal to zero.

r cos+ (a+ b)2 cos = 0 (2.71)

Hence, the angular velocity of the link 2 is

2 = r cos

(a+ b) cos(2.72)

The same result one can get by dierentiation of the absolute angular displacement = 3600 ) (see Fig. 24)

Y

X C

2 1 B

(t) G

A

r

3

d a

b

X

y

x

Figure 24

2 = = ddtarcsin

d+ r sina+ b

= r cos

(a+ b) cos(2.73)

The velocity of the point C, according to 2.70 is

vC = I (r sin+ (a+ b)2 sin) (2.74)

Dierentiation of the above expression with respect to time yields the wanted accel-eration of the point C.

aC = I

r sin r2 cos r cos tan + r2 sin tan r 1

cos2

(2.75)


Problem 20

G

l

Y

X

R

v

A

A

B

Figure 25

The end A of the rod shown in Fig. 25 slides over the cylindrical surfacewith the constant velocity vA. Derive the expression for the angular velocity andacceleration of the rod and the linear velocity and acceleration of its centre of gravityG.


Solution

G

l

Y

X

R

v

A

A

B

C

2 R

rG

Figure 26

Let us denote the absolute angular displacement of the rod by as shown inthe Fig. 26. It can be seen from this drawing that

bCA = 2R (2.76)

Since the point A moves with the constant velocity vAbCA = vAt (2.77)

Hence,

=vAt2R

(2.78)

Dierentiation of the above relationship yields the absolute angular velocity

= =vA2R

(2.79)

The absolute angular acceleration is

= = 0 (2.80)

The absolute linear velocity of the point G can be obtained by dierentiation of theabsolute position vector rG.

rG = I(l cosR cos 2) + J(l sin+ (RR sin 2)) (2.81)

Hence the absolute velocity of the point G is

vG = rG = I(l sin+R2 sin 2) + J(l cosR2 cos 2)) (2.82)

the absolute acceleration of the point G is

aG = vG = I(l sinl2 cos+ 2R sin 2+ 4R2 cos 2)+J(l cos l2 sin 2R cos 2+ 4R2 sin 2)) (2.83)


The absolute velocity of the point B which belong to the rod can be obtained by dif-ferentiation of its absolute position vector or from the following relationship betweentwo points which belong to the same body.

vB = vA + rBA

= IvA sin 2+ J(vA cos 2) +

I J K0 0 2R cos cos 2R cos sin 0

= I(vA sin 2 2R cos sin) + J(vA cos 2+ 2R cos cos)= I(vA sin 2 vA cos sin) + J(vA cos 2+ vA cos cos) == IvA cos sin+ JvA sin2 (2.84)

As one could predict, the absolute velocity of the point B possesses always directionparallel to the position of the rod

vBYvBX

=vA sin2

vA cos sin= tan (2.85)

and its scalar magnitude

vB = vA

q(cos sin)2 + sin4 = vA sin (2.86)

is equal to the component of the velocity of the point A along the rod


Problem 21

Y

X

C

a

2

1

A

B

1 1

O

a 2

b 2 2,1

Figure 27

The two elements 1 and 2 are joint together to form the double pendulumshown in Fig. 27. Motion of the element 1 is given by the angular displacement 1(t).The relative motion of the element 2 with respect to the element 1 is determined bythe angular displacement 2,1(t).

Produce:1. the expression for the absolute angular velocity and acceleration of the link 2

Answer:2 = K(1 + 21) 2 = K(1 + 21)2. the linear velocity and acceleration of the point CAnswer:vC = I(a11 sin1 a2(1 + 21) sin(1 +21) b2(1 + 21) cos(1 +21))++J(a11 cos1 + a2(1 + 21) cos(1 +21) b2(1 + 21) sin(1 +21))aC = I(a11 sin1a121 cos1a2(1+21) sin(1+21)a2(1+21)2 cos(1+21) b2(1 + 21) cos(1 +21) + b2(1 + 21)2 sin(1 +21))++J(a11 cos1 a121 sin1+ a2(1+ 21) cos(1+21) a2(1+ 21)2 sin(1+21) b2(1 + 21) sin(1 +21) b2(1 + 21)2 cos(1 +21))


Problem 22

Y

X

C 2

1

B

(t)

G

A

D

A X (t)

b a

Figure 28

The link 1 of the system shown in Fig. 28 moves along axis X of the inertialsystem of coordinates XY and its motion is given by the function of time XA(t). Thelink 2 of the rectangular shape is hinged to the link 1 at the point A. Its relativemotion with respect to the link 1 is given by the function of time (t).

Produce:1. the expression for the absolute angular acceleration of the link 2Answer:2 = K 2 = K2. the expression for the absolute linear velocity and linear acceleration of the pointG.Answer:vG = I(XA b2 sin

a2 cos) + J( b

2 cos a

2 sin)

aG = I(XA b2 sinb22 cos a

2 cos + a

22 sin) + J( b

2 cos b

22 sin

a2 sin a

22 cos)


Problem 23

A

B

D

l

l

Y

X

1 2 3

E

v l

C

Figure 29

The link 1 of the system shown in Fig. 29 can be treated as a rigid body. Thislink is being raised by means of the rope 2, the pulley 3. The radius of the pulley isnegligible. The point E of the rope moves with the constant velocity v.

Produce:1. the angular displacement of the link 1 as a function of timeAnswer: = 90o 2 arcsin

2lvt2l

2. the angular velocity of the link 1 as a function of timeAnswer: = v

l cos(45o2 )= 1l

vt14l2 (2l

2t2v2+2ltv2)

3. the absolute velocity of the point CAnswer:vC = 2l

PLANE KINETICS OF A RIGID BODY 86

2.2 PLANE KINETICS OF A RIGID BODY

2.2.1 Kinetic energyRotation of rigid body

Let us consider a rigid body that rotates with an angular velocity about the axisZ of the absolute system of coordinates XY Z (see Fig. 30).The kinetic energy of a

Y

X

x

y

Z,z

r

dm

rD

DO

rA Oo

Figure 30

particle of mass dm is

dT =1

2v2dm (2.87)

where v represents the absolute velocity of the particle. As it has been shown in theprevious chapter, magnitude of the absolute velocity v is

v= |rA| = | r| (2.88)where is the absolute angular velocity of the body and r is the vector that representsthe distance of the particle from the axis of rotation. Since these two vectors areperpendicular to each other and dm = dV

dT =1

22r2dV (2.89)

Hence, the total kinetic energy is the following volume integral

T =1

22ZZZ

Vr2dV

(2.90)


The expression in brackets is called (mass) moment of inertia and is usually denotedby I. Since in our case the body rotates about axis z through the point O we willdenote it by IzO. As a matter of convenience the above volume integral will bedenoted by

Rm.

IzO =ZZZ

Vr2dV =

Zm

r2dm (2.91)

General motion.

Y

X O

G

x

Z

r

dm

D

y

z

rG

rA r

DG

Figure 31

Kinetic energy of a particle dm of a rigid body (see Fig. 31) is

dT =1

2r2Adm (2.92)

where rA is the absolute position vector of the particle dm. If the body performs thegeneral motion, the position vector rA may be defined as follows

rA= rG+rD,G + r (2.93)

The vector rG determines position of the centre of gravity G with respect to theinertial space XY Z. The vector rD,G = KrD,G determines the plane of motion ofthe particle dm with respect to the centre of gravity. Since the magnitude of rD,Gdoes not depend on time and K is unit vector associated with the absolute systemof coordinates, its derivative is equal to 0. Hence, dierentiation of the above vectoryields

rA= rG + r


Introducing Eq. 2.93 into Eq. 2.92 one can obtain that

dT =1

2(rG + r)

2dm =1

2(r2Gdm+ r

2dm+ 2rG rdm) (2.94)

Integration over the entire body yields the total kinetic energy in the fallowing form

T =1

2(r2G

Zm

dm+Zm

r2dm+ 2rG Zm

rdm) (2.95)

It is easy to see that the first integral yields the entire mass of the bodyZm

dm = m (2.96)

Let us consider the second integralZm

r2dm =Zm

( r)2dm = 2Zm

r2dm = 2IzG (2.97)

In the last expression IzG stands for the (mass) moment of inertia of the body aboutaxis z through the centre of gravity G.

Let us consider the last integralZm

rdm =Zm

rdm = Zm

rdm (2.98)

On the other hand, the distance between the centre of gravity and the axis z isdetermine by the following formula

ro,G =1

m

Zm

rdm (2.99)

Since in the case consider the system of coordinates was chosen through the centreof gravity G, ro,G is equal to zero. It follows that the integral

Rmrdm must be equal

to zero too. Hence Zm

rdm = 0 (2.100)

Implementation of Eqs. 2.96, 2.97 and 2.100 into Eq. 2.95 gives the following formulafor kinetic energy.

T =1

2v2Gm+

1

22IzG (2.101)

where:vG is the absolute velocity of the centre of gravity of the body is the absolute angular velocity of the bodym is mass of the body


IzG is mass moment of inertia about axis through centre of gravity and per-pendicular to the plane of motion

The last formula permits to formulate the following statement.

STATEMENT: Kinetic energy of a rigid body is equal to a sum of itsenergy in the translational motion with velocity of its centre of gravity(12v2Gm energy of translation) and energy in the rotational motion about axis

through its centre of gravity (122 IzG energy of rotation).

2.2.2 Properties of the mass moment of inertiaThe introduced definition of the mass moment of inertia about axis z

Iz =Zm

r2dm (2.102)

allows us to calculate it for bodies having a simple geometrical shape like a cylinder,sphere, rectangular block etc. As an example let as develop formula for the momentof inertia of a cylinder of the radius R, the length L and the density (. (see Fig.32).According to the introduced definition, the mass moment of inertia of the cylinder

x

y

z

r

L

odr

R

Figure 32

of the radius r and the thickness of its wall dr is

dIz = r2dm = r2dAL( = r22rdrL( = 2L(r3dr

Hence the moment of inertia of the whole cylinder is

Iz =Z R0

2L(r3dr = 2L(Z R0

r3dr = 2L(

r4

4

R0

= L(R4

2= R2L(

R2

2= m

R2

2


where m is total mass of the cylinderIn a similar manner the formulae for dierent solids can be derived. Results

of such derivations are collected in the appendix B.For bodies having more complicated shape (see Fig. 33) a division into small

elements has to be carried out. Then, each element can be considered as a particle

x

y

z

o

mi

ir

Figure 33

and the integration may be replaced by summation.

Iz NXi=1

r2imi (2.103)

Parallel axis theorem

Let us assume, that the moment of inertia Iz of a body is known about the axis zof the xyz system of coordinates (Fig. 34). Let a, b be coordinates of the centre ofgravity G of the body considered. Let xG,yG,zG be the body system of coordinatesparallel to xyz having its origin at G.Moment of inertia of the body about the axis


O

G

ydm

xG

yG

r G

rd

x

xG

yG

a

b

Figure 34

z, according to the previously introduced definition, is

Iz =Zm

r2dm =Zm

((a+ xG)2 + (b+ yG)2)dm =

=

Zm

(a2 + x2G + 2axG + b2 + y2G + 2byG)dm

=

Zm

(x2G + y2G)dm+

Zm

(a2 + b2)dm+Zm

2axGdm+Zm

2byGdm

=

Zm

r2Gdm+Zm

d2dm+Zm

2axGdm+Zm

2byGdm

= IzG + d2

Zm

dm+ 2aZm

xGdm+ 2bZm

yGdm (2.104)

But,1

m

Zm

xGdm,1

m

Zm

yGdm (2.105)

represents components of the distance between origin G and the centre of gravity G,which is actually 0. Hence,

Iz = IzG + d2m (2.106)

The last formula is known as the parallel axes theorem.2.2.3 Equations of motionGeneral motion of rigid body

Let us consider a rigid body that performs the general motion with respect to theinertial space XY Z (see Fig 35).


Y

X O

G

x

Z

r

dm

D

y

z

rG

rA ri

DG

Fi

i

j=1

J F ij

Figure 35

This motion is uniquely determined by the position vector rG and the angularvelocity . Let us split the rigid body into individual particles the body is made of.Motion of one particle dmi is governed by the Newtons law. It says that the productof mass of the particle and the vector of its absolute acceleration is equal to the vectorof the resultant force acting on the particle. The forces that act on the particle canbe divided into two categories external and internal. The external forces are due tothe interaction of this particle with the external sources such as the Earth (gravityforces) or another elements that do not belong to the body considered. Let us denotethe resultant force due to the external forces by Fi. The internal forces are due tothe interaction of the particle i with all the other particles the body is made of. Theresultant of all the internal forces is denoted here by

PJj=1Fij. Hence, application of

the Newtons second law to the particle i yields

dmirA = Fi +JXj=1

Fij (2.107)

In the above equation rA stands for the absolute position vector. According to theFig. 35 it can be expressed as follows

rA = rG + rDG + ri (2.108)

Two subsequent dierentiations with respect to time produce the wanted absoluteacceleration of the particle i.

rA= rG + ri (2.109)rA = rG + ri+ ri (2.110)


Introduction of the above expression into equation 2.107 yields

dmirG + dmi ri+dmi ri= Fi +JXj=1

Fij (2.111)

It is possible to create such an equation for each particle the body is made of. Then,we can add these equation together getting

IXi=1

dmirG +IXi=1

dmi r+IXi=1

dmi r =IXi=1

Fi +IXi=1

JXj=1

Fij (2.112)

Let us consider each term of equation 2.112 separately.

IXi=1

dmirG = rGIXi=1

dmi = mrG (2.113)

I=1Xi

dmi ri= I=1Xi

dmiri (2.114)

The expression 1mPI=1

i dmiri represents oset of the centre of gravity of the bodyfrom the axis z. In the case considered this distance is equal to 0. Hence

I=1Xi

dmi ri= 0 (2.115)

For the same reason the next term is equal to zero too.

IXi=1

dmi ri= IXi=1

dmiri= IXi=1

dmi ri= IXi=1

dmiri= 0 (2.116)

PIi=1Fi represents the resultant of all external forces acting on the rigid body and it

will be denoted here by F.IXi=1

Fi = F (2.117)

Since the particle i can not interacts with itself,.Fii in the last term of equation 2.112must be equal to zero. Therefore, the double sum

PIi=1

PJj=1Fij is assembled of the

following elementsFij + Fji

which, according to the Newtons third law (Fij = Fji), are equal to zero. Hence

IXi=1

JXj=1

Fij = 0 (2.118)


Introducing equations from 2.113 to 2.118 into equation 2.112 one can obtain

mrG = F (2.119)

Another independent equation we can get by multiplying left and right hand side ofequation 2.132 by the vector r.

r (dmirG + dmi ri+dmi ri) = ri(Fi +JXj=1

Fij) (2.120)

or

dmirirG + dmiri( ri)+dmiri( ri) = riFi + riJXj=1

Fij (2.121)

Let us consider the triple cross-product existing in the above equation.

ri( ri) = (riri) ri(ri) (2.122)Taking into account that = k, riri=r2i , and since the vectors ri and are per-pendicularthat ri = 0 the above triple cross-product is

ri( ri) = kr2i (2.123)Similarly

ri( ri) = (riri) ri(ri) = (riri) = (( ri) ri))= ((riri) ) = 0 (2.124)

Introduction of 2.123 and 2.124 into 2.121 yields

dmirirG + dmikr2i= riFi + riJXj=1

Fij (2.125)

If one sum up such equations for all particles the body is made of, we can have

IXi=1

dmirirG +IXi=1

dmikr2i=IXi=1

riFi +IXi=1

(riJXj=1

Fij) (2.126)

ButIXi=1

dmikr2i = k

IXi=1

r2i dmi = kIzG (2.127)

Taking into account that 1mPI=1

i dmiri= 0

IXi=1

dmirirG = (IXi=1

dmiri) rG = 0 (2.128)


Since r represents position vector of the particle i and therefore is independent fromindex of summation j,

IXi=1

(riJXj=1

Fij) =IXi=1

JXj=1

(riFij) (2.129)

The double sum contains only elements riFij + rjFji. ButriFij + rjFji = riFij rjFij = (ri rj)Fij = rjiFij = 0 (2.130)

The above expression must be equal to zero since the two vectors rji and Fijareparallel (see Fig. 36) Introduction of the relationships from 2.127 to 2.130 into 2.126

G

y

r

x

ji

jr ri

dmi

dmj

FijFji

Figure 36

yields

kIzG =IXi=1

riFi (2.131)

Multiplying the above expression by the unit vector k one can get

IzG =IXi=1

k (riFi) =MzG (2.132)

The equation 2.132 together with equation 2.119 forms set of equation that is knownas the generalized Newtons equations

mrG = FIzG =MzG (2.133)

The first equation is equivalent to two scalar equations which can be produced bymultiplying it by two unit vectors associated with arbitrarily chosen system of coor-dinates xy. If one assume that the velocity of the centre of gravity is resolved alongthe body system of coordinates,

vG = rG = i(vGx) + j(vGy) (2.134)


the acceleration a = rG is

rG = v0G + vG = i(vGx) + j(vGy) +

i j k0 0 vGx vGy 0

=

= i(vGx vGy) + j(vGy + vGx) (2.135)

krodkiewski dynamic book

Documents

plane dynamics

motion of rigid body

rigid body592

plane kinematics

dynamics of plane mechanisms

plane kinetics

plane mechanical systems

rigid bodiesin terms